New York Journal of Mathematics
New York J. Math. 11(2005)635–647.
Tiling systems and homology of lattices in tree products
Guyan Robertson
Abstract. Let Γ be a torsion-free cocompact lattice in Aut(T1)×Aut(T2), whereT1,T2are trees whose vertices all have degree at least three. The group H2(Γ,Z) is determined explicitly in terms of an associated 2-dimensional tiling system. It follows that under appropriate conditions the crossed productC∗- algebraAassociated with the action of Γ on the boundary ofT1× T2 satisfies rankK0(A) = 2·rankH2(Γ,Z).
Contents
1. Introduction 635
2. Products of trees and their automorphisms 637
3. Some related graphs 640
4. Tilings andH2(Γ,Z) 641
5. K-theory of the boundaryC∗-algebra 645
5.1. One vertex complexes 645
5.2. Irreducible lattices in PGL2(Qp)×PGL2(Q) 646
References 647
1. Introduction
This article is motivated by the problem of calculating the K-theory of certain crossed product C∗-algebrasA(Γ, ∂Δ), where Γ is a higher rank lattice acting on an affine building Δ with boundary ∂Δ. Here we examine the case where Δ is a product of trees. We determine the K-theory rationally, thereby proving some conjectures in [KR].
LetT1 andT2be locally finite trees whose vertices all have degree at least three.
Consider the direct product Δ =T1× T2 as a two-dimensional cell complex. Let Γ be a discrete subgroup of Aut(T1)×Aut(T2) which acts freely and cocompactly on Δ. Associated with the action (Γ,Δ) is a tiling system whose set of tiles is the set R of “directed” 2-cells of Γ\Δ. There are vertical and horizontal adjacency rules
Received October 16, 2005, and in revised form on November 22, 2005.
2000Mathematics Subject Classification. 22E40, 22D25.
Key words and phrases. tree products, lattices, homology, K-theory, operator algebra.
ISSN 1076-9803/05
635
tHsandtV s between tilest, s∈Rillustrated in Figure1. Precise definitions will be given in Section 2.
t s t
s
Figure1.
There are homomorphismsT1, T2:ZR→ZR defined by T1t=
tHs
s, T2t=
tV s
s .
Consider the homomorphismZR→ZR⊕ZRgiven by T1−I
T2−I
:t→(T1t−t)⊕(T2t−t).
The main result of this article is the following theorem, which is formulated more precisely in Theorem 4.1.
Theorem 1.1. There is an isomorphism
H2(Γ,Z)∼= kerT1−I
T2−I
. (1)
The proof of (1) is elementary, but care is needed because the right-hand side is defined in terms of “directed” 2-cells rather than geometric 2-cells. A square complexX is VH-T if every vertex link is a complete bipartite graph and if there is a partition of the set of edges into vertical and horizontal, which agrees with the bipartition of the graph on every link [BM]. The universal covering space Δ of a VH-T complexX is a product of trees T1× T2 and the fundamental group Γ ofX is a subgroup of Aut(T1)×Aut(T2) which acts freely and cocompactly onT1× T2. Conversely, every finite VH-T complex arises in this way from a free cocompact action of a group Γ on a product of trees. Recall that a discrete group which acts freely on a CAT(0) space is necessarily torsion-free.
The group Γ acts on the (maximal) boundary ∂Δ of Δ, which is the set of chambers of the spherical building at infinity, endowed with an appropriate topology [KR]. This boundary may be identified with a direct product of Gromov boundaries
∂T1×∂T2. The boundary action (Γ, ∂Δ) gives rise to a crossed productC∗-algebra A(Γ, ∂Δ) =CC(∂Δ)Γ as described in [KR].
Ifpis prime then PGL2(Qp) acts on its Bruhat–Tits treeTp+1, which is a homo- geneous tree of degreep+ 1. Ifp, are prime then the group PGL2(Qp)×PGL2(Q) acts on Δ =Tp+1× T+1. Let Γ be a torsion-free irreducible lattice in PGL2(Qp)× PGL2(Q). Then A(Γ, ∂Δ) is a higher rank Cuntz–Krieger algebra and fits into the general theory developed in [RS1, RS2]. In particular, it is classified up to isomorphism by its K-theory. It is a consequence of Theorem 1.1 (see Section 5) that
rankK0(A(Γ, ∂Δ)) = 2·rankH2(Γ,Z).
(2)
This proves a conjecture in [KR]. The normal subgroup theorem [Mar, IV, The- orem (4.9)] implies thatH1(Γ,Z) is a finite group. Equation (2) can therefore be expressed as
χ(Γ) = 1 +1
2rankK0(A(Γ, ∂Δ)).
One easily calculates that χ(Γ) = (p−1)(4−1)|X0|, where |X0| is the number of vertices ofX. Therefore the rank ofK0(A(Γ, ∂Δ)) can be expressed explicitly in terms ofp, and|X0|. Examples are constructed in [M3, Section 3], wherep, ≡1 (mod 4) are two distinct primes.
2. Products of trees and their automorphisms
IfT is a tree, there is a type mapτ defined on the vertex set ofT, taking values inZ/2Z. Two vertices have the same type if and only if the distance between them is even. Any automorphism g of T preserves distances between vertices, and so there existsi∈Z/2Z(depending ong) such thatτ(gv) =τ(v) +i, for every vertex v.
Suppose that Δ is the 2-dimensional cell complex associated with a product T1× T2 of trees. Let Δk denote the set of k-cells in Δ fork= 0,1,2. The 0-cells are vertices and the 2-cells are geometric squares. Denote byu= (u1, u2) a generic vertex of Δ. There is a type mapτ on Δ0defined by
τ(u) = (τ(u1), τ(u2))∈Z/2Z×Z/2Z.
Any 2-cellδ∈Δ2has one vertex of each type. For every g∈AutT1×AutT2 there exists (k, l)∈Z/2Z×Z/2Zsuch that, for each vertexu,
τ(gu) = (τ(u1) +k, τ(u2) +l).
(3)
Let Γ<AutT1×AutT2 be a torsion-free discrete group acting cocompactly on Δ.
ThenX = Γ\Δ is a finite cell complex with universal covering Δ. LetXk denote the set ofk-cells of X fork= 0,1,2.
•
•
•
• 10
01
00
11
Figure 2. The model squareσ.
The first step is to formalize the notion of a directed square inX. We modify the terminology of [BM, Section 1], in order to fit with [RS1, RS2, KR]. Letσbe a model typed square with vertices00,01,10,11, as illustrated in Figure 2. Assume that the vertexijofσhas type
τ(ij) = (i, j)∈Z/2Z×Z/2Z.
The vertical and horizontal reflectionsv, hofσare the involutions satisfyingv(00) = 01, v(10) =11, h(00) =10, h(01) =11. An isometryr:σ→Δ is said to betype rotating if there exists (k, l)∈Z/2Z×Z/2Zsuch that, for each vertexijofσ
τ(r(ij)) = (i+k, j+l).
LetR denote the set of type rotating isometriesr:σ→Δ. Ifg∈AutT1×AutT2
and r ∈ R then it follows from (3) that g◦r ∈ R. If δ2 ∈ Δ2 then for each (k, l) ∈Z/2Z×Z/2Zthere is a unique r ∈R such thatr(σ) =δ2 and r(00) has type (k, l). Therefore each geometric square δ2 ∈ Δ2 is the image of each of the four elements of{r∈R ; r(σ) =δ2} under the mapr→ r(σ). The next lemma records this observation.
Lemma 2.1. The map r→r(σ)fromR toΔ2 is4-to-1.
Let R = Γ\R and call R the set of directed squares of X = Γ\Δ. There is a commutative diagram
R −−−−−→r→r(σ) Δ2
⏐⏐
⏐⏐ R −−−−→η X2
where the vertical arrows represent quotient maps and η is defined by η(Γr) = Γ·r(σ). The next result makes precise the fact that each geometric square inX2 corresponds to exactly four directed squares.
Lemma 2.2. The map η:R→X2 is surjective and 4-to-1.
Proof. Fixδ2∈R. By Lemma2.1, the set
{r∈R; r(σ) =δ2}={r1, r2, r3, r4} contains precisely 4 elements. Since Γ acts freely on Δ, the set
{Γr1,Γr2,Γr3,Γr4} ⊂R
also contains precisely four elements, each of which maps to Γδ2 under η. Now suppose thatη(Γr) = Γδ2for some r∈R. Thenγr(σ) =δ2 for some γ∈Γ. Thus γr∈ {r1, r2, r3, r4} and Γr∈ {Γr1,Γr2,Γr3,Γr4}. This proves thatη is 4-to-1.
The vertical and horizontal reflections v, h of the model square σ act on R and generate a group Σ ∼= Z/2Z×Z/2Z of symmetries of R. The Σ-orbit of each r ∈ R contains four elements. Choose once and for all a subset R+ ⊂ R containing precisely one element from each Σ-orbit. The map η restricts to a 1-1 correspondence between R+ and the set of geometric squares X2. For each φ∈Σ− {1}, letRφ denote the image of R+ under φ. Then Rmay be expressed as a disjoint union
R=R+∪Rv∪Rh∪Rvh.
Now we formalize the notion of horizontal and vertical directed edges inX. Con- sider the two directed edges [00,10],[00,01] of the model square σ.
[00,01]
[00,10].... ...........................
...
...
Figure 3. Directed edges of the model squareσ.
LetA be the set of type rotating isometriesr: [00,10]→Δ, and letB be the set of type rotating isometriesr: [00,01]→Δ. There is a natural 2-to-1 mapping r→ranger, fromA∪B onto Δ1. LetA= Γ\AandB= Γ\B. CallA,Bthe sets of horizontal and verticaldirected edges ofX = Γ\Δ. LetE =A∪B, the set of all directed edges ofX.
Ifa= Γr∈A, leto(a) = Γr(00)∈X0 andt(a) = Γr(10)∈X0, theorigin and terminus of the directed edgea. Similarly, if b= Γr∈B, leto(b) = Γr(00)∈X0 andt(b) = Γr(01)∈X0. Note that it is possible thato(e) =t(e).
A straightforward analogue of Lemma2.2 shows that each geometric edge inX1 is the image of each of two directed edges. The horizontal and vertical reflections onσinduce an inversion onE, denoted bye→e, with the property thate=eand o(e) =t(e). The pair (E, X0) is thus a graph in the sense of [Se]. Choose once and for all an orientation of this graph: that is a subset E+ of E, with E =E+ E+. WriteA+ =A∩ E+ and B+ =B∩ E+. The images ofA[respectively B] in X1 are the edges of thehorizontal[vertical] 1-skeleton Xh1 [Xv1].
Lemma 2.3. There is a well-defined injective map t→(a(t), b(t)) :R→A×B which is surjective ifX has one vertex.
b(t) a(t)
t
...........
.... ............
....
...
...
Figure 4. Directed edges inX.
Proof. The mapr→(r|[00,10], r|[00,01]) :R→A×Bis injective because each geo- metric square of Δ is uniquely determined by any two edges containing a common vertex.
Ift= Γr∈Rthen define
a(t) = Γr|[00,10], b(t) = Γr|[00,01].
Using the fact that Γ acts freely on Δ it is easy to see that the mapt→(a(t), b(t)) is injective.
IfX has one vertex, then any two elementsa∈A,b∈Bare represented by type rotating isometriesr1: [00,10]→Δ,r2: [00,01]→Δ withr1(00) =r2(00). The isometries r1, r2 are restrictions of an isometry r ∈ R, which defines an element
t= Γr∈Rwitha=a(t) andb=b(t).
Ift= Γr∈R, define directed edgesa(t)∈A, b(t)∈Bopposite toa(t), b(t), as follows.
a(t) = Γ(r◦v|[00,10]), b(t) = Γ(r◦h|[00,01]).
b(t) b(t) a(t)
a(t) t
...........
.... ............
....
...........
.... ............
... ....
... ......
Figure 5. Opposite edges.
In other words
a(t) =a(tv); b(t) =b(th).
(4)
3. Some related graphs
Associated to the VH-T complexX are two graphs (in the sense of [Se]) whose vertices are directed edges of X. Denote by Gv(A) the graph whose vertex set is A and whose edge set is R, with origin and terminus maps defined by t → a(t) andt→a(t) respectively. SimilarlyGh(B) is the graph whose vertex set isBand whose edge set is R, with the origin and terminus maps defined by t → b(t) and t→b(t).
a(t)
a(t) t
...........
.... ............
....
...........
.... ............
....
b(t) t b(t)
...
...
...
...
Figure 6. Edges ofGv(A) andGh(B).
Now define two directed graphs whose vertices are elements of R. The “hori- zontal” graphGh(R) has vertex set R. A directed edge [t, s] is defined as follows.
Consider the model rectangle H made up of two adjacent squares with vertices {(i, j)∈Z2 :i= 0,1,2, j= 0,1} where the vertex (i, j) has type (i+ 2Z, j+ 2Z).
The model squareσof Figure2 is considered as the left-hand square of H.
σ
Figure 7. The model rectangleH.
An isometry r : H → Δ is said to be type rotating if there exists (k, l) ∈ Z/2Z×Z/2Zsuch that, for each vertex (i, j) of H, τ(r((i, j))) = (i+k, j+l). A directed edge of Gh(R) is Γr where r : H → Δ is a type rotating isometry. The origin of Γr is t= Γr1, where r1 =r|σ and the terminus of Γr is s= Γr2, where r2 : σ → Δ is defined by r2(i, j) = r(i+ 1, j). There is a similar definition for
the “vertical” graphGv(R) with vertex setR. Edges [t, s] ofGh(R) andGv(R) are illustrated in Figure 8, by the ranges of representative isometries. These directed graphs are not graphs in the sense of [Se]: the existence of a directed edge [t, s]
does not in general imply the existence of a directed edge [s, t].
An edge ofGh(R) An edge ofGv(R)
t s t
s
Figure 8
Since Γ acts freely on Δ, it is easy to see that the existence of a directed edge [t, s] ofGh(R) with origint∈Rand terminuss∈Ris equivalent to
b(s) =b(t), s=th. (5)
Similarly the existence of a directed edge [t, s] of Gv(R), with origin t ∈ R and terminuss∈R is equivalent to
a(s) =a(t), s=tv. (6)
The next lemma will be used later. Recall that a lattice Γ in PGL2(Qp)×PGL2(Q) is automatically cocompact [Mar, IX Proposition 3.7)].
Lemma 3.1. If Γis a torsion-free irreducible lattice inPGL2(Qp)×PGL2(Q) (p andprime)acting on the corresponding product of trees, then the directed graphs Gh(R),Gv(R)are connected.
Proof. This follows from [M3, Proposition 2.15], using the topological transitivity of an associated shift system. The proof uses the Howe–Moore theorem forp-adic semisimple groups and is explained in [M2, Lemma 2].
4. Tilings and H
2(Γ , Z )
Throughout this section,T1andT2are locally finite trees whose vertices all have degree at least three. The group Γ acts freely and cocompactly on the 2-dimensional cell complex Δ = T1× T2 and we continue to use the notation introduced in the preceding sections.
For t, s∈ R write tHs [respectively tV s] to mean that there is a “horizontal”
[respectively “vertical”] directed edge [t, s] inGh(R) [respectively Gv(R)]. Define homomorphismsT1, T2:ZR→ZRby
T1t=
tHs
s, T2t=
tV s
s.
It follows from (5), (6) that T1t=
⎛
⎝
b(s)=b(t)
s
⎞
⎠−th,
T2t=
⎛
⎝
a(s)=a(t)
s
⎞
⎠−tv.
Consider the homomorphism T1−I
T2−I
: ZR→ZR⊕ZR,
t→(T1t−t)⊕(T2t−t).
Defineε:ZE →ZE+ by
ε(x) =
x ifx∈ E+,
−x ifx∈ E+. The boundary map∂:ZR+→ZE+ is defined by
∂t=ε(a(t) +b(t)−a(t)−b(t))
and sinceX is 2-dimensional,H2(Γ,Z) = ker∂. Define a homomorphism ϕ2:ZR+→ZR
by
ϕ2t=t−tv−th+tvh.
The rest of this section is devoted to proving the following result, which is a more precise version of Theorem 1.1.
Theorem 4.1. The homomorphismϕ2 restricts to an isomorphism fromH2(Γ,Z) ontokerT1−I
T2−I
.
Define a homomorphismϕ1:ZE →ZR⊕ZR by ϕ1(a) = 0⊕
⎛
⎝
a(s)=a
s−
a(s)=a
s
⎞
⎠, ifa∈A,
ϕ1(b) =
⎛
⎝
b(s)=b
s−
b(s)=b
s
⎞
⎠⊕0, ifb∈B.
Note that ifx∈ E thenϕ1(x) =−ϕ1(x) and soϕ1(ε(x)) =ϕ1(x).
Lemma 4.2. The homomorphisms ϕ1, ϕ2 are injective and the following diagram commutes:
ZE+ ←−−−−∂ ZR+
ϕ1⏐⏐ ⏐⏐ϕ2
ZR⊕ZR ←−−−−−
T1−I T2−I
ZR.
(7)
Proof. Lett∈R. Then
(T1−I)t=
⎛
⎝
b(s)=b(t)
s
⎞
⎠−th−t,
(T1−I)tv=
⎛
⎝
b(s)=b(t)
s
⎞
⎠−tvh−tv,
(T1−I)th=
⎛
⎝
b(s)=b(t)
s
⎞
⎠−t−th,
(T1−I)tvh=
⎛
⎝
b(s)=b(t)
s
⎞
⎠−tv−tvh. Therefore
(T1−I)◦ϕ2(t) = (T1−I)(t−tv−th+tvh)
=
⎛
⎝
b(s)=b(t)
s−
b(s)=b(t)
s
⎞
⎠−
⎛
⎝
b(s)=b(t)
s−
b(s)=b(t)
s
⎞
⎠. By definition ofϕ1, this implies that
ϕ1(b(t)−b(t)) = (T1−I)ϕ2(t)⊕0.
Similarly
ϕ1(a(t)−a(t)) = 0⊕(T2−I)ϕ2(t).
Therefore
T1−I
T2−I
◦ϕ2(t) =ϕ1(b(t)−b(t) +a(t)−a(t))
=ϕ1◦ε(b(t)−b(t) +a(t)−a(t))
=ϕ1◦∂(t).
This shows that (7) commutes.
It is obvious that ϕ2 is injective. To verify that ϕ1 is injective, define ψ : ZR⊕ZR→ZE+byψ(s, t) =ε(b(s)−a(t)). Then ψ◦ϕ1(x) is a nonzero multiple ofx, for allx∈ E. It follows thatψ◦ϕ1:ZE+ →ZE+is injective and therefore so
isϕ1.
Lemma 4.3. The homomorphism ϕ2 restricts to an isomorphism from H2(Γ,Z) ontoϕ2(ZR+)∩kerT
1−I
T2−I
. Proof. Letϕ2(β)∈kerT
1−I
T2−I
, whereβ ∈ZR+. It follows from (7) that ϕ1◦∂(β) = 0.
Butϕ1 is injective, so∂β= 0, i.e.,β∈H2(Γ,Z).
Conversely, ifβ ∈H2(Γ,Z) thenT1−I
T2−I
◦ϕ2(β) = 0 by (7), so ϕ2(β)∈kerT1−I
T2−I
.
Sinceϕ2is injective, the conclusion follows.
The next result, combined with Lemma4.3, completes the proof of Theorem 4.1.
Lemma 4.4. There is an inclusionkerT
1−I
T2−I
⊂ϕ2(ZR+).
Proof. Letα=
t∈Rλ(t)t ∈kerT1−I
T2−I
. We show thatα∈ϕ2(ZR+). Ifs∈R then the coefficient ofsin the sum representing (T1−I)αis
⎛
⎜⎜
⎜⎝
t∈R,t=sh b(t)=b(s)
λ(t)
⎞
⎟⎟
⎟⎠−λ(s) =
⎛
⎜⎜
⎝
t∈R b(t)=b(s)
λ(t)
⎞
⎟⎟
⎠−λ(s)−λ(sh).
This coefficient is zero, sinceα∈ker(T1−I). Therefore λ(s) +λ(sh) =
t∈R b(t)=b(s)
λ(t).
(8)
The right-hand side of Equation (8) depends only on b(s), so for any b ∈ B we define
μ(b) =
t∈R b(t)=b
λ(t).
Thus (8) may be rewritten as
λ(s) +λ(sh) =μ(b(s)).
(9)
It follows from (8) and (4) that
μ(b(s)) =μ(b(sh)) =μ(b(s)).
(10)
b(s) s b(s)
...
... ......
Figure 9. μ(b(s)) =μ(b(s))
Fix an element b0 ∈ B, and let C be the connected component of the graph Gh(B) containing b0. Then C is a connected graph with vertex set C0 ⊂ B and edge setC1⊂R. The graphChas a natural orientationC+=C1∩(R+∪Rv) and it is clear thatC1=C+∪ {th:t∈ C+}. Each vertex ofC has degree at least three, since the same is true of the tree T1. Therefore the number of vertices of C is less than the number of geometric edges, i.e.,|C0|<|C+|.
If b∈ C0 then there is a path inC0 from b0 to b. It follows by induction from (10) thatμ(b0) =μ(b). Thus
μ(b0) =
t∈R b(t)=b
λ(t) =
t∈C1 b(t)=b
λ(t).
Therefore
|C0|μ(b0) =
b∈C0
t∈C1 b(t)=b
λ(t) =
t∈C1
λ(t)
=
t∈C+
(λ(t) +λ(th)) =
t∈C+
μ(b(t))
=
t∈C+
μ(b0) =|C+|μ(b0).
Since|C0|<|C+|, it follows thatμ(b0) = 0 for allb0∈B. In other words, by (9), λ(s) =−λ(sh)
(11)
for all s ∈ R. A similar argument, using α ∈ker(T2−I) and interchanging the roles of horizontal and vertical reflections, shows that
λ(s) =−λ(sv) (12)
for alls∈R. Combining (11) and (12) gives λ(s) =λ(svh) (13)
for alls∈R. Finally,
α=
t∈R+
λ(s)s+λ(sv)sv+λ(sh)sh+λ(svh)svh
=
t∈R+
λ(s)
s−sv−sh+svh
=
t∈R+
λ(s)ϕ2(s)∈ϕ2(ZR+).
5. K-theory of the boundary C
∗-algebra
The (maximal) boundary ∂Δ of Δ is defined in [KR]. It is homeomorphic to
∂T1×∂T2, where∂Tj is the totally disconnected space of ends of the treeTj. The group Γ acts on∂Δ and hence onCC(∂Δ) viag→αg, whereαgf(ω) =f(g−1ω), for f ∈CC(∂Δ),g∈Γ. The full crossed productC∗-algebraA(Γ, ∂Δ) =CC(∂Δ)Γ is the completion of the algebraic crossed product in an appropriate norm. We present examples where the rank of the analyticK-groupK0(A(Γ, ∂Δ)) is determined by Theorem 4.1.
5.1. One vertex complexes. The case where the quotient VH-T complexXhas one vertex was studied in [KR]. The group Γ acts freely and transitively on the vertices of Δ and A(Γ, ∂Δ) is isomorphic to a rank-2 Cuntz–Krieger algebra, as described in [RS1, RS2]. The proof of this fact given in [KR, Theorem 5.1]. It follows from [RS1] that A(Γ, ∂Δ) is classified by its K-theory. By the proofs of [RS2, Proposition 4.13] and [KR, Lemma 4.3, Theorem 5.3], we have
K0(A(Γ, ∂Δ)) =K1(A(Γ, ∂Δ)) and
rank(K0(A(Γ, ∂Δ))) = 2·dim kerT1−I
T2−I
.
Together with Theorem4.1, this proves
rankK0(A(Γ, ∂Δ)) = 2·rankH2(Γ,Z).
(14)
This verifies a conjecture in [KR].
5.2. Irreducible lattices in PGL2(Qp)×PGL2(Q). Ifp, are prime then the group PGL2(Qp)×PGL2(Q) acts on the Δ =Tp+1×T+1and on its boundary∂Δ, which can be identified with a direct product of projective linesP1(Qp)×P1(Q).
Let Γ be a torsion-free irreducible lattice in PGL2(Qp)×PGL2(Q). Then Γ acts freely on Δ andA(Γ, ∂Δ) is a rank-2 Cuntz–Krieger algebra in the sense of [RS1].
The irreducibility condition (H2) of [RS1] follows from Lemma 3.1. The proofs of the remaining conditions of [RS1] are exactly the same as in [KR, Lemma 4.1].
It follows that (14) is also true in this case. Since Γ is irreducible, the normal subgroup theorem [Mar, IV, Theorem (4.9)] implies that H1(Γ,Z) = Γ/[Γ,Γ] is finite. Equation (14) can therefore be written
χ(Γ) = 1 +1
2rankK0(A(Γ, ∂Δ)).
(15)
On the other hand, one easily calculates
χ(Γ) = (p−1)(−1) 4 |X0|
where |X0| is the number of vertices of X. Therefore the rank ofK0(A(Γ, ∂Δ)) can be expressed explicitly in terms ofp, and|X0|.
Explicit examples are studied in [M3, Section 3]. Ifp, l≡1 (mod 4) are two dis- tinct primes, Mozes constructs an irreducible lattice Γp,in PGL2(Qp)×PGL2(Ql) which acts freely and transitively on the vertex set of Δ. Here is how Γp,l is con- structed. Let H(Z) = {a = a0+a1i+a2j+a3k;aj ∈ Z}, the ring of integer quaternions, letip be a square root of−1 inQp and define
ψ:H(Z)→PGL2(Qp)×PGL2(Q) by
ψ(a) =
a0+a1ip a2+a3ip
−a2+a3ip a0−a1ip
,
a0+a1i a2+a3i
−a2+a3i a0−a1i
.
LetΓp,={a∈H(Z);a0≡1 (mod 2), aj ≡0 (mod 2), j = 1,2,3,|a|2=prls}. Then Γp, = ψ(Γp,). The fact that Γp, is irreducible follows easily from [RR, Corollary 2.3], where it is observed that the only nontrivial direct product subgroup of Γp, isZ×Z=Z2.
Since|X0|= 1, it follows from (15) that
rankK0(A(Γ, ∂Δ)) = (p−1)(−1)
2 −2.
This proves an experimental observation of [KR, Example 6.2]. The construction of Mozes has been generalized in [Rat, Chapter 3] to all pairs (p, l) of distinct odd primes and the same conclusion applies.
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School of Mathematics and Statistics, University of Newcastle, NE1 7RU, U.K.
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